Evaluate the double integral. $ \int_0^{2\pi} \int_{\cos(x)}^{\cos(x) + 1} x + y \, dy \, dx =$ Choose 1 answer: Choose 1 answer: (Choice A) A $\pi + 2\pi^2$ (Choice B) B $0$ (Choice C) C $4\pi + 2$ (Choice D) D $\pi + 2$
Explanation: First, we evaluate the inner integral. We can substitute in the $\cos(x)$ and $\cos(x) + 1$ at the end as if they were numerical bounds. $\begin{aligned} & \int_0^{2\pi} \int_{\cos(x)}^{\cos(x) + 1} x + y \, dy \, dx \\ \\ &= \int_0^{2\pi} \left[ xy + \dfrac{y^2}{2} \right]_{\cos(x)}^{\cos(x) + 1} \, dx \\ \\ &= \int_0^{2\pi} x(\cos(x) + 1) + \dfrac{(\cos(x) + 1)^2}{2} \\ \\ &- \left( x\cos(x) + \dfrac{\cos^2(x)}{2} \right) \, dx \\ \\ &= \int_0^{2\pi} x + \cos(x) + \dfrac{1}{2} \, dx \end{aligned}$ Second, we evaluate the outer integral. $\begin{aligned} \int_0^{2\pi} x + \cos(x) + \dfrac{1}{2} \, dx &= \dfrac{x^2}{2} + \sin(x) + \dfrac{x}{2} \bigg|_0^{2\pi} \\ \\ &= \dfrac{4\pi^2}{2} + \sin(2\pi) + \dfrac{2\pi}{2} \\ \\ &- \left( \dfrac{0}{2} + \sin(0) + \dfrac{0}{2} \right) \\ \\ &= \pi + 2\pi^2 \end{aligned}$ The answer: $ \int_0^{2\pi} \int_{\cos(x)}^{\cos(x) + 1} x + y \, dy \, dx = \pi + 2\pi^2$